DFS_Algo_3
Context: This question pertains to the use of the Graph Abstract Data Type (ADT) implemented with an Adjacency Map, as studied in our course.
In the provided DFS algorithm: what is the time complexity of DFS for a graph represented using an adjacency map, where V is the number of vertices and E is the number of edges?
def DFS(g, u, discovered): for e in g.incident_edges(u): v = e.opposite(u) if v not in discovered: discovered[v] = e # mark v as discovered via edge e DFS(g, v, discovered)
Refer to the provided Graph ADT information for context.

class Vertex: def __init__(self, x): self._element = x
class Edge: def __init__(self, u, v, x): self._origin = u self._destination = v self._element = x
def opposite(self, v): return self._destination if v == self._origin else self._origin
class Graph: def __init__(self, directed=False): self._outgoing = {} self._incoming = {} if directed else self._outgoing
def incident_edges(self, v): return self._outgoing[v].values()

Question

DFS_Algo_3
Context: This question pertains to the use of the Graph Abstract Data Type (ADT) implemented with an Adjacency Map, as studied in our course.
In the provided DFS algorithm: what is the time complexity of DFS for a graph represented using an adjacency map, where V is the number of vertices and E is the number of edges?
def DFS(g, u, discovered):    for e in g.incident_edges(u):        v = e.opposite(u)        if v not in discovered:            discovered[v] = e  # mark v as discovered via edge e            DFS(g, v, discovered)
Refer to the provided Graph ADT information for context.

class Vertex:    def __init__(self, x):        self._element = x
class Edge:    def __init__(self, u, v, x):        self._origin = u        self._destination = v        self._element = x
    def opposite(self, v):        return self._destination if v == self._origin else self._origin
class Graph:    def __init__(self, directed=False):        self._outgoing = {}        self._incoming = {} if directed else self._outgoing
    def incident_edges(self, v):        return self._outgoing[v].values()

BlackTom AI · Accepted Answer

Question restatement: We are analyzing a DFS implementation on a graph represented with an adjacency map. The function DFS(g, u, discovered) iterates over for e in g.incident_edges(u), computes v = e.opposite(u), and if v is not yet discovered, marks it and recursively calls DFS(g, v, discovered). The graph API provided uses an adjacency map and the Graph.incident_edges(v) returns the collection of outgoing edges from v. The options for time complexity are: O(V + E), O((V + E) log V), O(E), O(V * E), O(V).

Option 1: O(V + E)
- This is the classic time complexity for DFS on a graph using an adjacency list or adjacency map representation. Each vertex is visited once, and every edge is explored (via incident_edges) at most twice (once from each endpoint in undirected graphs, or once along a directed edge). The aggregate work across all recursive calls sums to a linear function in the number of vertices and edges. Since the adjacency map access and edge iteration are the dominant costs, this aligns with the standard DFS cost model.

Option 2: O((V + E) log V)
- This would imply an extra logarithmic factor, typically arising if we used a priority queue or a balanced tree to order processing or if some operation costs log V. In the presented DFS, there is no sorting, priority selection, or log-cost operation per edge or per vertex. The iteration over incident_edges and the constant-time checks for discovery do not introduce a log factor, so this option overestimates the cost.

Option 3: O(E)
- This would neglect the cost related to visiting isolated vertices and the initial checks for each vertex. DFS must also account for the time to discover and traverse edges incident to each vertex; especially for graphs where V is large relative to E (e.g., sparse graphs), missing the V term would be inaccurate. Therefore, E alone is insufficient as a bound.

Option 4: O(V * E)
- This suggests a nested loop pattern where for each vertex we inspect all edges, leading to a quadratic-like interaction in terms of V and E. In the DFS as implemented, edges are not re-scanned from scratch for every vertex; once an edge is traversed, it doesn’t cause a full repetition across all vertices. The recursion and edge exploration happen in a controlled, near-linear fashion rather than a full multiplication of V and E.

Option 5: O(V)
- This would imply linear growth solely with the number of vertices, ignoring how edges contribute to work. In a graph with many edges (dense graphs), DFS must process edges and their connections, so E cannot be ignored. Therefore, O(V) is too small to bound the running time in general.

Summary of analysis across options:
- The correct characterization of DFS on a graph with adjacency map is dominated by visiting each vertex once and examining each edge at most a constant number of times, yielding a total time proportional to V + E.
- The other options introduce extra factors or omit essential contributions from edges, making them incorrect for the given DFS implementation with an adjacency map.

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